My textbook states we need locally Lipschitz continuous with respect to the second argument and uniformly with respect to the first argument of f(t, x). This is said to be equivalent to the result that for all compact subsets V⊂U where U⊆Rn+1 is open we haveL=⊃(t,x)≠(t,y)∈V⊂U|f(t,x)−f(t,y)||x−y|&lt;∞I understand that locally Lipschitz means there is a neighborhood around each point where f is Lipschitz, but what does the uniform part mean with respect to t?

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My textbook states we need locally Lipschitz continuous with respect to the second argument and uniformly with respect to the first argument of f(t, x). This is said to be equivalent to the result that for all compact subsets V⊂U where U⊆Rn+1 is open we haveL=⊃(t,x)≠(t,y)∈V⊂U|f(t,x)−f(t,y)||x−y|&amp;lt;∞I understand that locally Lipschitz means there is a neighborhood around each point where f is Lipschitz, but what does the uniform part mean with respect to t?

haillarip0c9 · Accepted Answer

Step 1From the first formulation you could get different constants for different times, so&amp;nbsp;L(t).In connecting the times in the Picard iteration and its bounds you want to use the same constant, so you need an uniform bound&amp;nbsp;L*&amp;nbsp;for&amp;nbsp;L(t).This bound itself is a constant for all the Lipschitz conditions, so you can also use that directly.More generally one only needs that&amp;nbsp;L(t)&amp;nbsp;is integrable to get a weight functionw(t)=exp⁡(−2|∫t0tL(s)ds|)instead of&amp;nbsp;exp⁡(−2L|t|)&amp;nbsp;for a norm that makes the Picard iteration contracting.

My textbook states we need locally Lipschitz continuous with respect to the second argument and unif

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